Question

A point on the ellipse $4x^2+9y^2=36$, where the normal is parallel to the line $4x-2y-5=0$, is?

• A. $(\frac{8}{5},\frac{-9}{5})$
• B. $(\frac{-9}{5},\frac{8}{5})$
• C. $(\frac{9}{5},\frac{8}{5})$
• D. $(\frac{8}{5},\frac{9}{5})$

Answer 1

Answer: (C)

$(\frac{9}{5},\frac{8}{5})$

Let $P(x_1,y_1)$, be the point on curve $4x^2+9y^2=36$, where the normal is parallel to line $4x-2y=5$.
$\therefore \frac{dy}{dx}{|}_{p(x_1,y_1)}=\left(\frac{-4x_1}{9y_1}\right)$
Now, $\frac{9y_1}{4x_1}=2\Rightarrow 9y_1=8x_1$..$\left(1\right)$
Also, $4x_1^2+9y_1^2=36$ $\left(2\right)$
$\therefore$ On solving $\left(1\right)$ and $\left(2\right)$, we get
$P(x_1=\frac{9}{5},y_1=\frac{8}{5})$ or $(x_1=\frac{-9}{5},y_1=\frac{-8}{5})$.